Collatz Conjecture

Comment:
Theorem 4. Let

$$
a_{n}=\sum_{i=0}^{n} \gamma_{i} 2^{i}, \quad n>1000, \quad \gamma_{i} \in\{0,1\}
$$

then

$$
a_{4 n}<a_{n}
$$
    \begin{proof}
.  Let's introduce operators defined formulas

$$
P f=f / 2, \quad T f=3 f+1, Z f=3 f\\
$$
$$
T_{i} \in\{P, T\}, R_{i} \in\{Z, P\},
$$

Let's consider all possible scenarios of the behavior of the Syracuse sequence , the same possible scenarios can be written in the following form

$$
\begin{gathered}
a_{n+n}=T_{1} T_{2} \ldots . T_{n} a_{n} \\
 % a_{n+n}=R_{1} R_{2} \ldots . R_{n} a_{n}+A_{2n}
\end{gathered}
$$
We need to calculate an estimate for every 2n-th member of the Collatz sequence based on the number of applied operators P ,T,Z over n step.\\

$$
\begin{gathered}
a_{n+n}=T_{n} T_{n-1} \ldots . T_{1} a_{n} \\
 % a_{n+n}=R_{1} R_{2} \ldots . R_{n} a_{n}+A_{2n}
\end{gathered}
$$
Let $a_n $have m ones in binary representation, then
 count the number of applications of the Z operator in the following formula.
$$
m=\sum_{R_{i}=Z,\,i\le n} 1
$$
and count the number of applications of the P operator in the following formula.
$$
\sum_{R_{i}=P,\, i\le n} 1= m+n-m 
$$
Because, each application Z  is accompanied by an operator P and the number of applications of the operator P according to the zeros $a_n$ has a corresponding n-m
By rules of Collatz we have after $n$ steps

$$
a_{n+n}=3^{m} / 2^{n} a_{n}+T_{n} T_{n-1} \ldots . T_{1}  1=3^{m} / 2^{n} a_{n}+B_{n}
$$

$$
B_{n} \leq \sum_{j=1, m} 3^{j}/2^{j}<23^{m} / 2^{m} \leq 3(3/2)^{m} 2^{-n}a_{n}
$$
$$
B_{n}  \leq 3(3/4)^{m} 2^{m-n}a_{n}
$$
According to the last formula, we see that the growth of each member of the sequence depends on the number of units in the binary representation. Next, we will show that a large number of ones at the * step leads to an increase in the number of zeros in the binary representation according to the previous theorems. Where will the decrease in the following terms of the sequence follow
$$
\begin{gathered}
a_{2 n}=3^{m}\left(a_{n} * 2^{-n}+B_{n}\right)=\left(a_{n} * 2^{-n}+B_{n}\right) 3^{m} \\
a_{2 n}=\sum_{i=0}^{\left[\alpha_{1}\right]} \gamma_{i} 2^{i}, \quad \gamma_{i} \in\{0,1\}, \quad \alpha_{1}=m * \ln 3 / \ln 2+\ln \left(2^{-n} a_{n}\right)
\end{gathered}
$$
Repeating the reasoning of Theorem 3, let's consider the equation
$$
2^x=a_{2 n}=3^{m}a_{n} \cdot 2^{-n}+B_{n} =3^{m}  +3^{m}(a_{n}-2^{n} )\cdot 2^{-n}  +B_{n}
$$
$$
x\ln2= m\ln(3) +\ln\left(1+(a_{n}-2^{n})\cdot 2^{-n}+ B_n \cdot 3^{-m}\right)       
$$
We come to the following
$$
x\ln2= m\ln(3) + \ln(1+\theta) + O( \frac{1}{2^n\ln2})     
$$
$$
\{x\}= \left\{\frac{m\ln(3)}{\ln(2)}   +\frac{\ln(1+\theta)}{\ln2} + O( \frac{1}{2^n\ln2}) \right\}    
$$
From the last equation, in order to apply the results of Theorem 3, we need $\{x\} >\ln2 $.
To execute the last inequality, consider the following cases

\begin{enumerate}
\item    $\gamma_{n-1}=0,\gamma_{n-2}=0$
\item    $\gamma_{n-1}=1,\gamma_{n-2}=1$
\item    $\gamma_{n-1}=0,\gamma_{n-2}=1$
\item    $\gamma_{n-1}=1,\gamma_{n-2}=0$
\end{enumerate}
In addition, we accept the possibility of changing m $\Rightarrow m\pm 1$. The latter is possible by changing the number of applications of the Collatz rules or, in other words, by decreasing or increasing the elements of the sequence by one. As a result, we have

$$
\{x\}= \left\{\frac{(m \pm 1)\ln(3)}{\ln(2)}   +\frac{\ln(1+\theta)}{\ln2} + O( \frac{1}{2^n\ln2}) \right\}    
$$

Considering the following cases for $m=m, m=m-1, m=m+1$ depending on the behavior of $\gamma_{n-1},\gamma_{n-2}$ we have the following three variants for the fractional part of x
$$
\{x\}= \left\{\frac{(m \pm 1)\ln(3)}{\ln(2)}   +\frac{\ln(1+\theta)}{\ln2} +O( \frac{1}{2^n\ln2}) \right\}    
$$
$$
\{x\}= \left\{\frac{m \ln(3)}{\ln(2)}   +\frac{\ln(1+\theta)}{\ln2} +O( \frac{1}{2^n\ln2}) \right\}    
$$
Thus, depending on the behavior of $\gamma_{n-1},\gamma_{n-2}$ we can always choose an option where the fractional part of x will satisfy the conditions of Theorem 3.

Denoting 
$$
\begin{aligned}
& m^{*} \text { is the number of non-zero } \gamma_{i} \\
& l^{*} \text { is the number of zero } \gamma_{i}
\end{aligned}
$$
by theorem 3 we will have
$$
\begin{gathered}
m^{*} \leq\left[\alpha_{1}\right] / 2+5=[m \ln 3 / \ln 2] / 2+5 \\
l^{*} \geq\left[\alpha_{1} / 2-5=[m \ln 3 / \ln 2] / 2-5\right.
\end{gathered}
$$

After $3n$ steps of applying Collatz rules, we have
$$
a_{4n} = \frac{3^{m^{*} } }{2^{2n   }}a_{2n}+T_{3n} T_{3n-1} \ldots . T_{1}  1= \frac{3^{m^{*} } }{2^{2n   }} a_{2n}+B_{3n  }
$$

$$
a_{4n} = \frac{3^{m^{*} } }{2^{2n   }} a_{2n}+T_{3n} T_{3n-1} \ldots . T_{1}  1= 
 \frac{3^{m^{*} } }{2^{2n   }} \left( \frac{3^{m } }{2^{n   }} a_{n}+B_{n}\right)+B_{3n  }
$$

$$
a_{4n} =3^{m^{*}+m }  2^{-3n}  a_{n}+  3^{m^{*} }  2^{-2n}B_{n}+B_{3n  }
$$
$$
B_{3n  }=3^{m^{*}}2^{-2n} B_{n  }=3^{m^* +m} 2^{-2n-n} a_{n  }
$$

$$
a_{4n}   \le  q_{1} * a_{n}
$$
by definitions $ m^*, l^*,B_{n} $  we get
$$
q_{1}<1
$$
 Using  $n > 1000 \Rightarrow q_{1} < 1 \Rightarrow a_{4n  } < a_{n}$.
\end{proof}
Theorem 5. Let
$$
a_{n}=\sum_{i=0}^{n} \gamma_{i} 2^{i}, \quad n>1000, \quad \gamma_{i} \in\{0,1\}
$$
then for $a_{n}$ Collatz conjecture is true
Proof. Proof follows from theorem 1-4
\section*{6. Conclusions}
Our assertion proves that after $3n$ steps, a sequence with an initial binary length of $n$ arrives at a number strictly smaller than the initial one, from which the solution to the Collatz conjecture follows. This is because by applying this process $n$ times, we are guaranteed to arrive at 1.